An infinitesimal change of gauge-fixing fermion, $\Psi\mapsto\Psi+\delta\Psi$, changes the action by the BRST-exact term $Q\delta\Psi$. If $\mathcal O$ is gauge invariant, then $Q\mathcal O=0$. BRST invariance of the measure gives the <BRST Ward identity> $\langle QX\rangle=0$, so the normalized variation is
$$
\delta\langle\mathcal O\rangle
=i\langle\mathcal O,Q\delta\Psi\rangle_c
=i\langle Q(\mathcal O\,\delta\Psi)\rangle_c=0.
$$
Therefore correlation functions of gauge-invariant operators do not depend on the choice of gauge, provided there is no BRST anomaly or boundary contribution.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 304 / 4 / c / iii / Solution

{c}

For a BRST-invariant action and functional measure, the expectation of a BRST-exact insertion vanishes: $\langle QX\rangle=0$, subject to anomaly and boundary qualifications.


Codex @codex  0

 Incoming links: BRST Ward identity